Sign in

username:

password:



Not a member?

Search Online Books



Search tips

Free Online Books

Ads

Chapters

See Also

Embedded SystemsFPGAElectronics
Chapter Contents:

Search Physical Audio Signal Processing

  

Book Index | Global Index


Would you like to be notified by email when Julius Orion Smith III publishes a new entry into his blog?

  

Two Masses Connected by a Rod

Figure B.5: Two ideal point-masses $ m$ connected by an ideal, rigid, massless rod of length $ 2r$.
\includegraphics[width=1.5in]{eps/massrodmass}

As an introduction to the decomposition of rigid-body motion into translational and rotational components, consider the simple system shown in Fig.B.5. The excitation force densityB.15 $ f(t,x)$ can be applied anywhere between $ x=-r$ and $ x=r$ along the connecting rod. We will deliver a vertical impulse of momentum to the mass on the right, and show, among other observations, that the total kinetic energy is split equally into (1) the rotational kinetic energy about the center of mass, and (2) the translational kinetic energy of the total mass, treated as being located at the center of mass. This is accomplished by defining a new frame of reference (i.e., a moving coordinate system) that has its origin at the center of mass.

First, note that the driving-point impedance7.1) ``seen'' by the driving force $ f(t,x)dx$ varies as a function of $ x$. At $ x=0$, The excitation $ f(t,0)dx$ sees a ``point mass'' $ 2m$, and no rotation is excited by the force (by symmetry). At $ x=\pm R$, on the other hand, the excitation $ f(t,\pm R)dx$ only sees mass $ m$ at time 0, because the vertical motion of either point-mass initially only rotates the other point-mass via the massless connecting rod. Thus, an observation we can make right away is that the driving point impedance seen by $ f(t,x)$ depends on the striking point $ x$ and, away from $ x=0$, it depends on time $ t$ as well.

To avoid dealing with a time-varying driving-point impedance, we will use an impulsive force input at time $ t=0$. Since momentum is the time-integral of force ( $ f=ma=m\dot{v}\,\,\Rightarrow\,\,mv=\int f\,dt$), our excitation will be a unit momentum transferred to the two-mass system at time 0.



Subsections
Previous: Circular Cross-Section
Next: Striking the Rod in the Middle

Order a Hardcopy of Physical Audio Signal Processing


About the Author: Julius Orion Smith III
Julius Smith's background is in electrical engineering (BS Rice 1975, PhD Stanford 1983). He is presently Professor of Music and Associate Professor (by courtesy) of Electrical Engineering at Stanford's Center for Computer Research in Music and Acoustics (CCRMA), teaching courses and pursuing research related to signal processing applied to music and audio systems. See http://ccrma.stanford.edu/~jos/ for details.


Comments


No comments yet for this page


Add a Comment
You need to login before you can post a comment (best way to prevent spam). ( Not a member? )